Is there an ''algorithmic'' way to get that intersection of two quadrics $$x_1 y_1 -x_2 y_2 - z_1^2+z_2^2=0$$ and $$x_2 y_1 + x_1 y_2 -2z_1z_2=0$$ inside $\mathbb{R}P^5[x_1:x_2:y_1:y_2:z_1:z_2]$ is isomorphic $\mathbb{R}P^3?$ By ''algorithmic'' I mean similarily to statements like ''degree d curve in $\mathbb{C}P^2$ has genus (d-1)(d-2)/2.''
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$\begingroup$ your question seems to be fairly specific; in this particular case, you could probably classify all smooth interesctions of two quadrics (and smoothness can be checked explicitly if you have the definining equations, I believe). I am not sure whether one can rule out continuous moduli (i.e. is it true that in a family of smooth real projective varieties with one fiber isomorphic to $\mathbb{R}P^3$ all fibers are isomorphic to $\mathbb{R}P^3$). If one can do this, then the classification would consists of a finite number of cases, which could be checked explicitly. $\endgroup$– user137767Commented Apr 4, 2019 at 19:48
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1$\begingroup$ "isomorphic to RP^3" meaning "birational over R" or something different/stronger? (Note too that since you specify that you're working over the reals, even a single smooth quadric need not be rational because there might be no rational points!) $\endgroup$– Noam D. ElkiesCommented Apr 4, 2019 at 19:54
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$\begingroup$ Well, I've put ''isomorphic'' on purpose as I didn't want to specify it. I was hoping of, say, ''isomorphic as a projective variety'', though I essentially need it to be homeomorphic/diffeomorphic to $\mathbb{R} P^3.$ $\endgroup$– FilipCommented Apr 4, 2019 at 23:40
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$\begingroup$ If you need only diffeomorphism, there was some work by Thom or Sullivan on classification of smooth complete intersections, which may be relevant to your "algorithmic" question. I don't know if that holds over real numbers. $\endgroup$– user74900Commented Apr 5, 2019 at 6:30
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$\begingroup$ I guess you need at least that your intersection of quadrics contains a line (which I think is not always the case over the reals), otherwise I wouldn't know how to do that. In that case the map from $\mathbb{P}^3$ to $X$ your intersection should go like this. Take a $\mathbb{P}^3$ disjoint from the line $L$ and pick $p\in \mathbb{P}^3$. Then consider the $\mathbb{P}^2=:Q$ spanned by $L$ and p. If all goes well, each quadric intersects $Q$ in two degenerate conics $L+M$ and $L+N$ with $M$ and $N$ lines. The coordinates of $M\cap N$ are the image of $p$. This is just a rational map, btw. $\endgroup$– IMeasyCommented Apr 5, 2019 at 15:09
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